Question

Consider the beta distribution with parameters $(\mathrm{a},\mathrm{b})$. Show that

(a) when $\mathrm{a}>1$and $\mathrm{b}>1$, the density is unimodal (that is, it has a unique mode) with mode equal to $(\mathrm{a}-1)/(\mathrm{a}+\mathrm{b}-2)$

(b) when $\mathrm{a}\le 1$, $\mathrm{b}\le 1$, and $\mathrm{a}+\mathrm{b}<2$, the density is either unimodal with mode at $0$or $1$or U-shaped with modes at both$0$and$1$;

(c) when $\mathrm{a}=1=\mathrm{b}$, all points in $[0,1]$are modes.

Short answer

a.$\left[0,1\right]$is a maximum segment of $\mathrm{x}$is unimodal

b. It is zero or one mode.

c. It is zero and one mode.

Step-by-step solution

Explanation (part a)

a.

Density function,

$\mathrm{f}\left(\mathrm{x}\right)=\frac{1}{\mathrm{B}(\mathrm{a},\mathrm{b})}{\mathrm{x}}^{\mathrm{a}-1}(1-\mathrm{x}{)}^{\mathrm{b}-1}$

Maximum of function,

$\mathrm{g}\left(\mathrm{x}\right)=(\mathrm{a}-1)\mathrm{logx}+(\mathrm{b}-1)\log (1-\mathrm{x})$

First derivative is,

${\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)=\frac{\mathrm{a}-1}{\mathrm{x}}-\frac{\mathrm{b}-1}{1-\mathrm{x}}$

Second derivative is,

${\mathrm{g}}^{\text{'}\text{'}}\left(\mathrm{x}\right)=-\frac{\mathrm{a}-1}{{\mathrm{x}}^2}+\frac{\mathrm{b}-1}{(1-\mathrm{x}{)}^2}$

For

$\mathrm{a}>1$and $\mathrm{b}>1$derivative is equal to zero

So,

${\mathrm{g}}^{\text{'}}\left(\mathrm{x}\right)=0\iff \frac{\mathrm{a}-1}{\mathrm{x}}$

$=\frac{\mathrm{b}-1}{1-\mathrm{x}}\iff \mathrm{x}=\frac{\mathrm{a}-1}{\mathrm{a}+\mathrm{b}-2}$

$\underset{\mathrm{x}\to 0}{\lim }\mathrm{g}\left(\mathrm{x}\right)=\underset{\mathrm{x}\to 1}{\lim }\mathrm{g}\left(\mathrm{x}\right)=-\infty$

Explanation (part b)

The stationary point calculated in $(0,1)$is the lone contender for maximum in interval$(0,1)$in this situation (a).

However, we do have that.

${\mathrm{g}}^{\text{'}\text{'}}\left(\frac{\mathrm{a}-1}{\mathrm{a}+\mathrm{b}-2}\right)=-\frac{(\mathrm{a}-1{)}^3}{(\mathrm{a}+\mathrm{b}-2{)}^2}+\frac{(\mathrm{b}-1{)}^3}{(\mathrm{a}+\mathrm{b}-2{)}^2}\ge 0$

Explanation (part c)

c.

If $\mathrm{a}=\mathrm{b}=1$, $\mathrm{X}~\mathrm{Unif}(0,1)$is the result.

However, it is self-evident that every point between 0 and 1 is a mode.